= Solution
A Riemannian metric $g$ on a <complex manifold> is a Kähler metric when its complex-linear extension is a <Hermitian form> on each tangent space and its fundamental two-form
$$
\omega(u,v)=g(Ju,v)
$$
is closed. Equivalently, $\omega$ is a positive real closed $(1,1)$-form, making $X$ a <Kähler manifold>.
The <Lefschetz operator of a Kähler manifold> and its adjoint are
$$
L\alpha=\omega\wedge\alpha,
\qquad \Lambda=L^*.
$$
Because $d\omega=0$ and $\omega$ has type $(1,1)$, both $\partial\omega$ and $\bar\partial\omega$ vanish. The graded <Leibniz rule> therefore gives $[L,\partial]=[L,\bar\partial]=0$.
Writing formal adjoints with stars, define the three <Laplace-Beltrami operator>[Laplacians] by
$$
\Delta=dd^*+d^*d,
\qquad
\Delta_\partial=\partial\partial^*+\partial^*\partial,
\qquad
\Delta_{\bar\partial}=\bar\partial\bar\partial^*+\bar\partial^*\bar\partial.
$$
The supplied <Kähler identities> identity $[\Lambda,\partial]=i\bar\partial^*$ gives, by complex conjugation and taking adjoints,
$$
[\Lambda,\bar\partial]=-i\partial^*,
\qquad [L,\partial^*]=i\bar\partial,
\qquad [L,\bar\partial^*]=-i\partial.
$$
Expanding these commutators and using $\partial\bar\partial+\bar\partial\partial=0$ shows that the mixed terms in $\Delta$ vanish and that $\Delta_\partial=\Delta_{\bar\partial}$. Since $d=\partial+\bar\partial$, it follows that
$$
\Delta=\Delta_\partial+\Delta_{\bar\partial}=2\Delta_\partial=2\Delta_{\bar\partial}.
$$
The adjoint of $[L,\partial]=0$ gives $[\Lambda,\partial^*]=0$. Hence
$$
[\Lambda,\Delta_\partial]=i(\bar\partial^*\partial^*+\partial^*\bar\partial^*)=0,
$$
where the last equality is the adjoint of $\partial\bar\partial+\bar\partial\partial=0$. Thus $\Lambda$, and therefore every power $\Lambda^k$, commutes with $\Delta_\partial$. It follows that $\Lambda^k$ sends every $\partial$-harmonic $(p,q)$-form to a $\partial$-harmonic $(p-k,q-k)$-form whenever $0\leq k\leq\min(p,q)$.
Solved by gpt-5.6-sol high.
Back to article page
